Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10, lg = 21;
int n, m, q;
vector<int> G[maxn];
int c[maxn], h[maxn];
vector<int> vec[maxn][lg];
int up[maxn][lg];
vector<int> Class[maxn];
void merge(vector<int> &v, vector<int> &u, vector<int> &w) {
  for (int ui : u) v.push_back(ui);
  for (int wi : w) v.push_back(wi);
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  v.resize(min((int)v.size(), 10));
}
void dfs(const int &v, const int &par) {
  h[v] = h[par] + 1;
  up[v][0] = par;
  for (int ci : Class[par]) vec[v][0].push_back(ci);
  sort(vec[v][0].begin(), vec[v][0].end());
  vec[v][0].resize(min((int)vec[v][0].size(), 10));
  for (int i = 1; i < lg; i++)
    up[v][i] = up[up[v][i - 1]][i - 1],
    merge(vec[v][i], vec[v][i - 1], vec[up[v][i - 1]][i - 1]);
  for (int u : G[v])
    if (u != par) dfs(u, v);
}
int lift(int u, int p) {
  for (int i = lg - 1; i >= 0; i--)
    if ((((p) >> (i)) & 1)) u = up[u][i];
  return u;
}
int get(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  u = lift(u, h[u] - h[v]);
  if (u == v) return u;
  for (int i = lg - 1; i >= 0; i--)
    if (up[u][i] != up[v][i]) u = up[u][i], v = up[v][i];
  return up[u][0];
}
void get_vec(vector<int> &res, int lca, int u) {
  int p = h[u] - h[lca];
  for (int i = lg - 1; i >= 0; i--) {
    if ((((p) >> (i)) & 1)) {
      for (int vi : vec[u][i]) res.push_back(vi);
      u = up[u][i];
    }
  }
  sort(res.begin(), res.end());
  res.resize(unique(res.begin(), res.end()) - res.begin());
}
void pre() {
  for (int v = 1; v <= n; v++) {
    for (int ci : Class[v]) vec[v][0].push_back(ci);
    sort(vec[v][0].begin(), vec[v][0].end());
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  scanf("%d%d%d", &n, &m, &q);
  int u, v;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &c[i]);
    Class[c[i]].push_back(i);
  }
  pre();
  h[0] = -1;
  dfs(1, 0);
  int a, lca, sz;
  vector<int> res;
  while (q--) {
    scanf("%d%d%d", &u, &v, &a);
    lca = get(u, v);
    res.clear();
    if (lca == u && lca == v) {
      sz = min(a, (int)Class[v].size());
      printf("%d ", sz);
      for (int i = 0; i < sz; i++) printf("%d ", Class[v][i]);
      printf("\n");
      continue;
    }
    get_vec(res, lca, u), get_vec(res, lca, v);
    sz = min((int)res.size(), a);
    printf("%d ", sz);
    for (int i = 0; i < sz; i++) printf("%d ", res[i]);
    printf("\n");
  }
}